Count of N-bit binary numbers without leading zeros

Count of N-bit binary numbers without leading zeros

Last Updated : 10 Mar, 2022

Given an integer N, the task is to find the count of N-bit binary numbers without leading zeros.
Examples:

Input: N = 2
Output: 2
10 and 11 are the only possible binary numbers.
Input: N = 4
Output: 8

Approach: Since the numbers cannot have leading zeros so the left-most bit has to be set to 1. Now for the rest of the N - 1 bits, there are two choices they can either be set to 0 or 1. So, the count of possible numbers will be 2^{N - 1}.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of possible numbers
int count(int n)
{
    return pow(2, n - 1);
}

// Driver code
int main()
{
    int n = 4;

    cout << count(n);

    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
    
    // Function to return the count 
    // of possible numbers 
    static int count(int n) 
    { 
        return (int)Math.pow(2, n - 1); 
    } 
    
    // Driver code 
    public static void main (String[] args) 
    { 
        int n = 4; 
    
        System.out.println(count(n)); 
    } 
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the count
# of possible numbers
def count(n):
    return pow(2, n - 1)

# Driver code
n = 4

print(count(n))

# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
    
class GFG 
{
    // Function to return the count 
    // of possible numbers 
    static int count(int n) 
    { 
        return (int)Math.Pow(2, n - 1); 
    } 
    
    // Driver code 
    public static void Main (String[] args) 
    { 
        int n = 4; 
    
        Console.WriteLine(count(n)); 
    }
}

// This code is contributed by 29AjayKumar

JavaScript

<script>


// JavaScript implementation of the approach

// Function to return the count
// of possible numbers
function count(n)
{
    return Math.pow(2, n - 1);
}

// Driver code
var n = 4;
document.write(count(n));


</script>

Output:

Time Complexity: O(log n)

Auxiliary Space: O(1)

Count of N-bit binary numbers without leading zeros

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